Answer:
It could possibly cause damage to the person lighting the firework and would quicken the ignition time.
Explanation:
The purpose of the longer fuse is so the person can run back in enough time to reach safety. Shortening that fuse can cause serious danger. The role of the fuse is to delay ignition.
If the secondary fuse of an aerial firework is too short, the firework would explode prematurely, potentially presenting a danger to operators and spectators.
In the context of aerial fireworks, fuses are critical components that control the timing of explosions. If the secondary fuse in an aerial firework is too short, the firework would explode prematurely, potentially before it even leaves the ground or far below the intended altitude. The short fuse wouldn't provide the requisite delay, which is normally intended to allow the firework to reach its maximum altitude before exploding. This could pose safety risks to those operating the fireworks and the nearby spectators as well.
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The two sub-atomic particles present in the nucleus of an aluminum (Al) atom are: proton and neutrons.
Protons: Positively charged protons are present in the atom's nucleus. They have an approximate relative mass of 1 atomic mass unit (amu), a relative charge of 1, and a relative mass of 1.
Neutrons: The nucleus of an atom contains neutrons, which are neutral particles. They have a relative mass of roughly 1 atomic mass unit (amu) and no electric charge (their charge is zero).
The number of protons in an aluminum atom determines its atomic number, which is 13, and the number of neutrons can vary, resulting in various aluminum isotopes with various masses. Aluminum-27, which has a nucleus made up of 13 protons and 14 neutrons, is the most prevalent isotope of the metal.
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a pure substance because it is an element.
an impure substance because it is chemically combined.
an impure substance because it is a mixture.
Answer is: a pure substance because it is a compound.
Pure substance is made of only one type of atom or only one type of molecule, it has definite and constant composition with distinct chemical properties.
Emerald (beryl) is chemical compound (beryllium aluminium cyclosilicate) with the chemical formula Be₃Al₂(SiO₃)₆. Pure beryl is colorless.
An emerald can be described as an impure substance because it is a mixture. Therefore, option D is correct.
The chemical formula for emerald is Be₃Al₂(SiO₃)₆, indicating that it is a compound composed of beryllium (Be), aluminum (Al), and silicon (Si) atoms bonded to oxygen (O) atoms.
However, emeralds can also contain impurities or trace elements such as chromium (Cr) or vanadium (V), which give them their characteristic green color.
These impurities or trace elements are present in small amounts and are incorporated into the crystal structure of the emerald.
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Answer: The temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K
Explanation:
According to ideal gas equation:
P = pressure of gas = 2300 mm Hg = 3.02 atm (760mmHg=1atm)
V = Volume of gas = 15 L
n = number of moles = 0.6
R = gas constant =
T =temperature = ?
Thus the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K
1) The binding energy per nucleon is 7.59 MeV/nucleon and 2) The binding energy per nucleon is 8.39 MeV/nucleon.
1) The binding energy per nucleon is as follows:
m_proton + m_neutron + m_electron - m_U
= 92*1.007276466812 + 92*5.4857990943e-4 + 143*1.008664916 - (235.0439299)
= 1.9151 u
= 931.494061 * 1.9151 MeV
= 1784 MeV
= 1784/235 = 7.59 MeV/nucleon
2) The binding energy per nucleon is as follows:
m_proton + m_neutron + m_electron - m_Cs
= 55*1.007276466812 + 55*5.4857990943e-4 + 82*1.008664916 - (136.9070895)
= 1.2338 u
= 931.494061 * 1.2338 MeV
= 1149 MeV
=1145/137 = 8.39 MeV/nucleon
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3.00 g
5.56 g
8.40 g
Answer : The amount left of leutium-176 will be, 2.10 g
Solution :
First we have to calculate the rate constant, we use the formula :
Now we have to calculate the amount left of the sample.
Expression for rate law for first order kinetics is given by :
where,
k = rate constant =
t = decay time =
a = initial amount of the sample = 16.8 g
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
Therefore, the amount left of leutium-176 will be, 2.10 g
Answer:
A.
Explanation:
The Gibbs free energy change for this reaction at 298 K is approximately -54,484.8 J.
Calculating the Gibbs free energy change (ΔG) for a reaction at a given temperature using the Gibbs-Helmholtz equation,
ΔG = ΔH - TΔS
ΔG = (-144,600 J) - (298 K)(-301.2 J/K)
ΔG = -144,600 J + 90,115.2 J
ΔG = -54,484.8 J
ΔG = -54,484.8 J
Converting to kilojoules:
ΔG = -54,484.8 J / 1,000
= -54.48 kJ
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